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Julie Déserti, Jérémy Blanc
To cite this version:
Julie Déserti, Jérémy Blanc. DEGREE GROWTH OF BIRATIONAL MAPS OF THE PLANE. Annali della Scuola Normale Superiore di Pisa, 2015, 14, pp.507-533. �hal-03000447�
arXiv:1109.6810v3 [math.AG] 7 Jun 2012
November 17, 2018
ABSTRACT. This article studies the sequence of iterative degrees of a birational map of the plane. This sequence is known either to be bounded or to have a linear, quadratic or exponential growth.
The classification elements of infinite order with a bounded sequence of degrees is achieved, the case of elements of finite order being already known. The coefficients of the linear and quadratic growth are then described, and related to geometrical properties of the map. The dynamical number of base-points is also studied. Applications of our results are the description of embeddings of the Baumslag-Solitar groups and GL(2, Q)
into the Cremona group.
2010 Mathematics Subject Classification. — 14E07, 37F10, 32H50
CONTENTS
1. Introduction 1
2. Elliptic maps 4
3. On the growth of the number of base-points 8
4. Growth of Jonquières twists 12
5. Growth of Halphen twists 13
6. Applications 16
References 19
1. INTRODUCTION
A rational map of the complex projective plane P2= P2
Cinto itself is a map of the following type
φ: P299K P2, (x : y : z) 99K φ0(x, y, z) :φ1(x, y, z) :φ2(x, y, z)
,
where theφi’s are homogeneous polynomials of the same degree without common factor. The degree degφ
of φis by definition the degree of these polynomials. We will only consider birational maps, which are rational maps having an inverse, and denote by Bir(P2) the group of such maps, classically called Cremona
group.
We are interested in the behaviour of the sequencedegφk
k∈N. According to [13], the sequence is either
bounded or has a linear, quadratic or exponential growth. We will say thatφis (1) elliptic if the growth is bounded;
(2) a Jonquières twist if the growth is linear; (3) an Halphen twist if the growth is quadratic;
Both authors were supported by the Swiss National Science Foundation grant no PP00P2_128422 /1.
(4) hyperbolic if the growth is exponential.
This terminology is classical, and consistent with the natural action of Bir(P2) on an hyperbolic space of
infinite dimension, where Jonquières and Halphen twists are parabolic ([9, Theorem 3.6]). Recall that the first dynamical degree ofφ∈ Bir(P2) isλ(φ) = lim
k→+∞(degφ
k)1/k
∈ R. This is an invariant
of conjugation which allows to distinguish the first three cases (where λ(φ) = 1) from the last case (where
λ(φ) > 1). There are plenty of articles on hyperbolic elements and the possible values for the algebraic integer
λ(φ), we are here more interested in the growth of the first three cases.
The nature of the growth is invariant under conjugation, and induces geometric properties onφ, that we describe now.
An elementφ∈ Bir(P2) is elliptic if and only if it is conjugate to an automorphism g ∈ Aut(S) of a smooth projective rational surface S such that gnbelongs to the connected component Aut0(S) of Aut(S) for some
n> 0 (see [13, Theorem 0.2, Lemma 4.1]). Reading this description, one would expect to find examples where g has infinite order and does not belong to the connected component. We will remove this possibility and refine the result of [13] in Section 2, by showing that, up to conjugation, either g has finite order or S= P2. The complete classification of elements of finite order of Bir(P2) can be found in [8]; for elements
of infinite order, one has (Proposition2.3):
Theorem A. Ifφ∈ Bir(P2) is elliptic of infinite order, thenφis conjugate to an automorphism of P2, which restricts to one of the following automorphisms on some open subset isomorphic to C2:
(1) (x, y) 7→ (αx,βy), whereα,β∈ C∗, and where the kernel of the group homomorphism Z2→ C∗given by(i, j) 7→αiβjis generated by(k, 0) for some k ∈ Z;
(2) (x, y) 7→ (αx, y + 1), whereα∈ C∗.
The end of Section2is devoted to the description of the conjugacy classes of such maps (Proposition2.4) and their centralizers in the Cremona group (Lemmas2.7and2.8).
A birational mapφ∈ Bir(P2) has a finite number b(φ) of base-points (that may belong to P2or correspond
to infinitely near points). We will call the number µ(φ) = lim
k→+∞
b(φk)
k ,
the dynamical number of base-points of φ. In Section 3, we study the sequence {b(φk)}k∈N and deduce
some properties on the number µ(φ); let us state some of them. It is a non-negative integer invariant under
conjugation; it also allows us to give a characterisation of birational maps conjugate to an automorphism of a smooth projective rational surface (Proposition3.5):
Theorem B. Let S be a smooth projective surface; the birational mapφ∈ Bir(S) is conjugate to an
auto-morphism of a smooth projective surface if and only if µ(φ) = 0.
In the case whereφis a Jonquières twist, the number µ(φ) determines the degree growth ofφ. A Jonquières twist preserves an unique pencil of rational curves [13, Theorem 0.2]. The sequencedegφk k∈Ngrows as
αk for some constant α∈ R. The number α is not invariant under conjugation, but one can show that the minimal value is attained when the rational curves of the pencil are lines, and is an integer divided by 2. More precisely, one has (Proposition4.4):
(1) The set lim k→+∞ deg(ψφkψ−1) k ψ∈ Bir(P2)
admits a minimum, which is equal to µ(2φ)∈12N.
(2) There exists an integer a∈ N such that lim k→+∞ deg(φk) k = µ(φ) 2 · a 2.
Moreover, a= 1 if and only ifφpreserves a pencil of lines.
The case of Halphen twists is similar. An Halphen twist preserves an unique pencil of elliptic curves (see [14], [13, Theorem 0.2]). Any such pencil can be sent by a birational map onto a pencil of curves of degree 3n with 9 points of multiplicity n, called Halphen pencil. We obtain the following (Proposition5.1):
Theorem D. Letφ∈ Bir(P2) be an Halphen twist.
(1) The set lim k→+∞ deg(ψφkψ−1) k2 ψ∈ Bir(P2) admits a minimumκ(φ) ∈ Q>0.
(2) There exists an integer a≥ 3 such that lim k→+∞ deg(φk) k2 =κ(φ) · a2 9. Moreover, a= 3 if and only ifφpreserves an Halphen pencil.
An application of our results is the description of birational maps whose two distinct iterates are conjugate, the non-existence of embeddings of Baumslag-Solitar groups into the Cremona group and the description of the embeddings of GL(2, Q) into the Cremona group (§6):
Theorem E. Letφdenote a birational map of P2of infinite order. Assume thatφnandφmare conjugate and that|m| 6= |n|. Then,φis conjugate to an automorphism of C2of the form(x, y) 7→ (αx, y + 1), whereα∈ C∗
such thatαm+n= 1 orαm−n= 1.
In particular, ifφis conjugate toφnfor any positive integer n, thenφis conjugate to(x, y) 7→ (x,y + 1).
Theorem F. If|m|, |n|, 1 are distinct, there is no embedding of
BS(m, n) = hr, s|rsmr−1= sni, m, n ∈ Z, mn 6= 0
into the Cremona group.
Theorem G. Letρ: GL(2, Q) → Bir(P2) be an embedding. Up to conjugation by an element of Bir(P2),
there exists an odd integer k and an homomorphismχ: Q∗→ C∗(with respect to multiplication) such that
ρ a b c d = x·χ(ad − bc) (cy + d)k , ay+ b cy+ d , ∀ a b c d ∈ GL(2,Q).
Acknowledgements. The authors would like to thank Yves de Cornulier and Ivan Marin for interesting
discussions on group theory. Thanks also to Charles Favre for questions on elliptic maps of infinite order and to Serge Cantat for his remarks and references.
2. ELLIPTIC MAPS
2.1. Classification of elliptic maps of infinite order. Ifφis a birational map of P2such thatdegφk k
∈N
is bounded, it is conjugate to an automorphism g of a smooth rational surface S such that the action of g on Pic(S) is finite [13, Lemma 4.1]. If g has finite order, the possible conjugacy classes are completely classified in [8]. Here we deal with the case of elements of infinite order, classifying the possibilities and describing its centralizers in Bir(P2).
Proposition 2.1. Let g be an automorphism of a smooth rational surface S which has infinite order but has
a finite action on Pic(S). Then, there exists a birational morphism S → X where X is equal either to P2or to an Hirzebruch surface Fnfor n6= 1, which conjugates g to an automorphism of X.
Remark 2.2. The proof of this result follows from a study of possible minimal pairs, which is similar to the
one made in [7] for finite abelian subgroups of Bir(P2) (see [7, Lemmas 3.2, 6.1, 9.7]).
Proof. Contracting the possible sets of disjoint (−1)-curves on S which are invariant by g, we can assume that the action of g on S is minimal. The action of g on Pic(S) being of finite order, the process corresponds
to applying a G-Mori program, where G is a finite group acting on Pic(S) (we only look at parts of the Picard
group which are invariant). Then one of the following occurs ([16,15]): (1) Pic(S)ghas rank 1 and S is a del Pezzo surface;
(2) Pic(S)ghas rank 2, and there exist a conic bundleπ: S→ P1on S, together with an automorphism h
of P1such that h◦π=π◦ g.
We want to show that S is P2or an Hirzebruch surface Fnfor n6= 1, and exclude the other cases.
In the case where Pic(S)ghas rank 1, the fact that g has infinite order but finite action on Pic(S) implies
that the kernel of the group homomorphism Aut(S) → Aut(Pic(S)) is infinite. So S is a del Pezzo surface of degree(KS)2≥ 6. The surface cannot be F1otherwise the exceptional section would be invariant. Similarly,
it cannot be the unique del Pezzo surface of degree 7, which has exactly three(−1)-curves, forming a chain (one touches the two others, which are disjoint), because the curve of the middle (and also the union of the two others) would be invariant. The only possibilities are thus P2, P1× P1= F0, and the del Pezzo surface of
degree 6.
If S is the del Pezzo surface of degree 6, any element h∈ Aut(S) acting minimally on S has finite order [7, Lemma 9.7]. Let us recall the simple argument. The del Pezzo surface of degree 6 is isomorphic to
S=(x : y : z), (u : v : w) ∈ P2× P2|xu = yv = zw .
The projections π1, π2 on each factor are birational morphisms contracting three (−1)-curves on p1 =
(1 : 0 : 0), p2= (0 : 1 : 0) and p3= (0 : 0 : 1). The group Pic(S) is generated by the six (−1)-curves of S,
which are Ei= (π1)−1(pi) and Fi= (π2)−1(pi) for 1 ≤ i ≤ 3 and form an hexagon. In fact, the action on the
hexagon gives rise to an isomorphism Aut(S) ≃ (C∗)2⋊(Sym
3⋊ Z/2Z) = (C∗)2⋊ D6. The action of g on S
being minimal, g permutes cyclically the curves, and either g or g−1acts as(E1→ F2→ E3→ F1→ E2→ F3).
This implies that g or g−1is equal to
(x : y : z), (u : v : w)→ (αv :βw : u), (βy :αz :αβx),
We can now assume the existence of an invariant conic bundleπ: S→ P1. Ifπhas no singular fibre, then S is a Hirzebruch surface Fnand n6= 1 because of the minimality of the action. It remains to exclude the case
whereπhas at least one singular fibre. The minimality of the action on S implies that the two components of any singular fibre F (which are two(−1)-curves) are exchanged by a power gkof g, and in particular that the whole singular fibre is invariant by gk. Note that k a priori depends on F.
We now prove that gk does not act trivially on the basis of the conic bundle. If gk acts trivially on the basis of the fibration, the automorphism g2k acts trivially on Pic(S); taking a birational morphism S → Fn
which contracts a component in each singular fibre one conjugates g2k to an automorphism of F
n, which
fixes pointwise at least one section. The pull-back on S of this section intersects only one component in each singular fibre and its image by gk gives thus another section, also fixed by g2k. The action of gk on a general fibre ofπexchanges the two points of the two sections and hence has order 2: contradiction.
The action of gkon the basis is non-trivial and fixes the point of P1corresponding to the singular fibre; so the same holds for g (recall that the fixed points of an element of Aut(P1) and any of its non-trivial powers
are the same). This implies that F is invariant by g, so its two components are exchanged by it (and thus k is odd).
In particular, g exchanges the two components of any singular fibre. This implies that the number of singular fibres ofπis at most 2, so S is the blow-up of one or two points of an Hirzebruch surface.
The fact that the two components of at least one singular fibre are exchanged gives a symmetry on the sections, that will help us to determine S. Denote by−m the minimal self-intersection of a section ofπand let s be one section which realises this minimum. Contracting the components in the singular fibres which do not intersect s, one has a birational morphism S→ Fn. The image of s is a section with minimal
self-intersection, so m= n. If n = 0, then taking some section of F0 = P1× P1 of self-intersection 0 passing
through at least one blown-up point, its strict transform on S would be a section of negative self-intersection, which contradicts the minimality of s2, so m= n > 0. Let us denote by s′ the section g(s), which also has
self-intersection −m on S but self-intersection −m + r on Fm, where r is the number of singular fibres ofπ.
Because any section of Fmdistinct from the exceptional section has self-intersection≥ m, we get −m+r ≥ m,
so 2≥ r ≥ 2m, which implies that m = 1 and r = 2.
The surface S is thus the blow-up of two points on F1, not lying on the exceptional section and not on the
same fibre, so is a del Pezzo surface of degree 6. The fact that g acts minimally on S is impossible, as we
already observed.
Proposition 2.3. Letφbe a birational map of P2of infinite order, such thatdegφk k∈Nis bounded.
Thenφis conjugate to an automorphism of P2, which restricts to one of the following automorphisms on some open subset isomorphic to C2:
(1) (x, y) 7→ (αx,βy), whereα,β∈ C∗, and where the kernel of the group homomorphism Z2→ C∗given by(i, j) 7→αiβjis generated by(k, 0) for some k ∈ Z;
(2) (x, y) 7→ (αx, y + 1), whereα∈ C∗.
Proof. According to Proposition2.1the mapφis conjugate to an automorphism of a minimal surface S, equal to either P2or an Hirzebruch surface.
Suppose first that S= P2. Looking at the Jordan normal form, any automorphism of P2is conjugate to
– either(x : y : z) 7→ (αx :βy : z),
– or(x : y : z) 7→ (αx : y+ z : z),
– or(x : y : z) 7→ (x + y : y + z : z).
This latter automorphism is conjugate to(x : y : z) 7→ (x : y + z : z) in Bir(P2) (for instance by (x : y : z) 99K (xz −1
2y(y − z) : yz : z
2), as already observed in [6, Example 1]). It remains to study the case of diagonal
we associate to each diagonal automorphism ψ: (x : y : z) 7→ (αx :βy : z) the kernel ∆ψ of the following homomorphism of groups: δψ: Z2→ C∗, (i, j) 7→αiβj. For any M= a b c d
∈ GL(2,Z), we denote by M(ψ) the diagonal automorphism (x : y : z) 7→ αaβbx :αcβdy : z,
which is the conjugate ofψby the birational map(x, y) 99K xayb, xcyd(viewed in the chart z= 1). We can
check that
δM(ψ)=δψ◦ tM,
which implies that∆M(ψ)= tM−1(∆
ψ). We can always choose M (by a result on Smith’s normal form) such
that∆M(φ)is generated by k1e1and k1k2e2, where e1, e2are the canonical basis vectors of Z2, and k1, k2are
non-negative integers, and replace φwith M(φ), which is conjugate to it. Since φand M(φ) have infinite
order, we see that k2= 0, and get the assertion on the kernel stated in the proposition.
If S= F0= P1× P1, we can reduce to the case of P2by blowing-up a fixed point and contracting the strict
transform of the members of the two rulings passing through the point.
Suppose now that S= Fnfor n≥ 2. If g fixes a point of Fn which is not on the exceptional section, we
can blow-up the point and contract the strict transform of the fibre to go to Fn−1. We can thus assume that all
points of Fnfixed by g are on the exceptional section. The action of g on the basis of the fibration is, up to
conjugation, x7→αx or x7→ x+1 for someα∈ C∗. Removing the fibre at infinity and the exceptional section, we get C2, where the action of g is
– either(x, y) 7→ (αx,βy+ Q(x)),
– or(x, y) 7→ (x + 1,βy+ Q(x)),
where α, β∈ C∗ and Q is a polynomial of degree ≤ n. The action on the fibre at infinity is obtained by conjugating by(x, y) 99K 1 x, y xn .
In the first case, there is no fixed point on the fibre at infinity (except the point on the exceptional section) if and only ifβ=αn and deg Q= n. There is no fixed point on x = 0 if and only if Q(0) 6= 0 and β= 1.
This implies that αis a primitive k-th rooth of unity, where k is a divisor of n. Conjugating by(x, y +γxd)
we replace Q(x) with Q(x) +γ(αd− 1)xd, so we can assume that the coefficient of xd is trivial if d is not
a multiple of k, which means that Q(x) = P(xk) for some polynomial P ∈ C[x]. In particular, g is equal
to(x, y) 7→ (ξx, y + P(xk)) and is conjugate to (x, y) 7→ ξx, y + 1by(x, y) 99Kx, y P(xk)
.
In the second case, there is no fixed point on C2, and no point on the fibre at infinity if and only ifβ= 1
and deg Q= n. Conjugating g by (x, y) 7→ (x,y +γxn+1) (which corresponds to performing an elementary link Fn99K Fn+1at the unique fixed-point and then coming back with an elementary link at a general point of the fibre at infinity), we get
(x, y) 7→ x + 1,y −γxn+1+ Q(x) +γ(x + 1)n+1.
Choosing the right elementγ∈ C, we can decrease the degree of Q(x), and get (x,y) 7→ (x+1,y) by induction.
2.2. Conjugacy classes of elliptic maps of infinite order. Following [6], we will call elements of the form
(x, y) 7→ (αx,βy), resp. (x, y) 7→ (αx, y + 1) diagonal automorphisms, resp. almost-diagonal automorphisms
Proposition 2.4 ([6], Theorem 1). (1) A diagonal automorphism and an almost-diagonal automorphism of C2are never conjugate in Bir(C2).
(2) Two diagonal automorphisms(x, y) 7→ (αx,βy) and (x, y) 7→ (γx,δy) are conjugate in Bir(C2) if and
only if there exists
a b c d
∈ GL(2,Z) such that (αaβb,αcβd) = (γ,δ).
(3) Two almost diagonal automorphisms (x, y) 7→ (αx, y + 1) and (x, y) 7→ (γx, y + 1) are conjugate
in Bir(C2) if and only ifα=γ±1.
Corollary 2.5. Letφ∈ Bir(P2) be an elliptic map which has infinite order. Ifφmis conjugate toφnin Bir(P2)
for some m, n∈ Z, |m| 6= |n|, thenφis conjugate to an automorphism of C2of the form(x, y) 7→ (αx, y + 1),
whereα∈ C∗such thatαm+n= 1 orαm−n= 1.
Proof. Note that mn6= 0 sinceφhas infinite order. Thenφis conjugate to one of the two cases of Proposi-tion2.3.
First of all, assume that up to conjugation φis(x, y) 7→ (αx,βy) and that the kernel ∆φ of the group ho-momorphism Z2→ C∗ given by (i, j) 7→αiβj is generated by(k, 0) for some k ∈ Z. Sinceφm and φn are conjugate there exists a matrix
N= a b c d ∈ GL(2,Z)
such that(αm)a(βm)b, (αm)c(βm)d= (αn,βn) (Proposition2.4). This means that
αma−nβmb,αmcβmd−n= (1, 1),
so (ma − n,mb), (mc,md − n) belong to∆φ. In particular mb= md − n = 0, which implies that b = 0, so ad= ±1, which is impossible since m 6= ±n.
Assume now that φis conjugate to (x, y) 7→ (αx, y + 1) for someα in C∗. The fact that φm and φn are
conjugate implies thatαm+n= 1 orαm−n= 1 (Proposition2.4).
2.3. Centralisers of elliptic maps of infinite order. Ifφis a birational map of P2, we will denote by C(φ)
the centraliser ofφin Bir(P2):
C(φ) =ψ∈ Bir(P2) φψ=ψφ .
In the sequel, we describe the centralisers of elliptic maps of infinite order of Bir(P2). The results are
groups which contain the centralisers of some elements of PGL(2, C). We recall the following result, whose
proof is an easy exercise. Recall that PGL(2, C) is the group of automorphisms of P1, or equivalently the
group of Möbius transformations x99K axcx+d+b.
Lemma 2.6. For anyα∈ C∗, we have
n η∈ PGL(2,C) η(αx) =αη(x)o= PGL(2, C) if α= 1 {x 99Kγx±1|γ∈ C∗} if α= −1 {x 7→γx|γ∈ C∗} if α6= ±1
Lemma 2.7. Let us consider φ: (x, y) 7→ (αx,βy) whereα, βare in C∗, and where the kernel of the group homomorphism Z2→ C∗ given by(i, j) 7→αiβj is generated by(k, 0) for some k ∈ Z. Then the centraliser ofφin Bir(P2) is
Proof. Let ψ: (x, y) 99K (ψ1(x, y),ψ2(x, y)) be an element of C(φ). The fact that ψ commutes withφ is
equivalent to
(⋆) ψ1(αx,βy) =αψ1(x, y) and (⋄) ψ2(αx,βy) =βψ2(x, y).
Writingψi=QPii for i= 1, 2, where Pi, Qiare polynomials without common factors, we see that P1, P2, Q1, Q2
are eigenvectors of the linear automorphismφ∗of the C-vector space C[x, y] given byφ∗: f(x, y) 7→ f (αx,βy).
This means that each of the Pi, Qi is a product of a monomial in x, y with an element of C[xk]. Using (⋆)
and(⋄), we get the existence of R1, R2∈ C(x) such that
ψ1(x, y) = xR1(xk), ψ2(x, y) = yR2(xk).
The fact thatψis birational implies thatψ1(x, y) is an elementη(x) ∈ PGL(2,C); it satisfiesη(αx) =αη(x)
because of(⋆).
Lemma 2.8. Let us consider φ: (x, y) 7→ (αx, y +β) whereα,β∈ C∗. The centraliser of φin Bir(P2) is
equal to
C(φ) =(x, y) 99K (η(x), y + R(x)) η∈ PGL(2,C),η(αx) =αη(x), R ∈ C(x),R(αx) = R(x) .
Proof. Conjugating by(x, y) 7→ (x,βy), we can assume thatβ= 1.
Letψ:(x, y) 99K (ψ1(x, y),ψ2(x, y)) be a birational map of P2which commutes withφ. One has
(⋆)ψ1(αx, y + 1) =αψ1(x, y) and (⋄)ψ2(αx, y + 1) =ψ2(x, y) + 1.
Equality(⋆) implies thatψ1only depends on x (see [6, Lemma 2]). Thereforeψ1is an element of PGL(2, C)
which commutes with x7→αx. Equality(⋄) implies that
∂ψ2 ∂y (αx, y + 1) = ∂ψ2 ∂y (x, y) and ∂ψ2 ∂x (αx, y + 1) =α −1∂ψ2 ∂x (x, y), which again means that ∂ψ2
∂x (x, y) and
∂ψ2
∂y (x, y) only depend on x. The second component ofψ can thus be
written ay+ B(x), where a ∈ C∗, B∈ C(x). Replacing this form in (⋄), we get B(αx) = B(x) + 1 − a,
which implies that ∂∂Bx(αx) =α−1∂B
∂x(x), and thus that x∂ B
∂x(x) is invariant under x 7→αx.
Ifα is not a root of unity, this means that ∂∂Bx = c/x for some c ∈ C; since B is a rational function, one gets c= 0 and B is a constant (or equivalently an element such that B(αx) = B(x)). It implies moreover a = 1
and we are done.
Ifαis a primitive k-th root of unity, the fact thatψ:(x, y) 99K (η(x), ay + B(x)) commutes with
φk:(x, y) 99K (x,y + k)
yields a(y + k) + B(x) = ay + B(x) + k, so a = 1. We again get B(αx) = B(x).
3. ON THE GROWTH OF THE NUMBER OF BASE-POINTS
If S is a projective smooth surface, any elementφ∈ Bir(S) has a finite number of base-points, which can belong to S or be infinitely near. We denote by b(φ) the number of such points. We will call the number
µ(φ) = lim k→+∞
b(φk)
the dynamical number of base-points ofφ. Since b(φψ) ≤ b(φ) + b(ψ) for anyφ,ψ∈ Bir(S), we see that µ(φ)
is a non-negative real number. Moreover, b(φ−1) and b(φ) being always equal, we get µ(φk) = |k · µ(φ)| for
any k∈ Z.
In this section, we precise the properties of this number, and will in particular see that it is an integer. Ifφ∈ Bir(S) is a birational map, we will say that a (possibly infinitely near) base-point p ofφis a persistent base-point if there exists an integer N such that p is a base-point ofφk for any k≥ N but is not a base-point ofφ−kfor any k≥ N.
If p is a point of S or a point infinitely near, which is not a base-point ofφ∈ Bir(S), we define a pointφ•(p),
which will also be a point of S or a point infinitely near. For this, take a minimal resolution Z π2 ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ π1 ✁✁✁✁ ✁✁✁✁ S φ ❴ ❴ ❴// ❴ ❴ ❴ ❴ S,
whereπ1,π2are sequences of blow-ups. Because p is not a base-point ofφ, it corresponds, viaπ1, to a point
of Z or infinitely near. Usingπ2, we view this point on S, again maybe infinitely near, and call itφ•(p).
Remark 3.1. If p is not a base-point of φ∈ Bir(S) and φ(p) is not a base-point ofψ∈ Bir(S), we have (ψφ)•(p) =ψ•(φ•(p)). If p is a general point of S, thenφ•(p) =φ(p) ∈ S.
Example 3.2. If S= P2, p= (1 : 0 : 0) andφis the birational map(x : y : z) 99K (yz + x2: xz : z2), the point
φ•(p) is not equal to p =φ(p), but is infinitely near to it.
Using this definition, we put an equivalence class on the set of points that belong to S or are infinitely near, by saying that p is equivalent to q if there exists an integer k such that(φk)•(p) = q (this implies that p is not
a base-point ofφk and that q is not a base-point ofφ−k). The set of equivalence classes is the generalisation of the notion of set of orbits for birational maps.
Proposition 3.3. Letφbe a birational map of a smooth projective surface S. Denote by νthe number of equivalence classes of persistent base-points ofφ. Then, the set
n b(φk) −νk k ≥ 0o⊂ Z is bounded.
In particular, µ(φ) is an integer, equal toν.
Proof. Let us say that a base-point q is periodic if(φk)•(q) = q for some k 6= 0, or if q is a base-point ofφk
for any k∈ Z \ {0} (which implies that (φk)•(q) is never defined for k 6= 0). Let us denote by
P
the set ofperiodic base-points ofφand by b
P
the finite set of points equivalent to a point ofP
.The number of base-points of φ and φ−1 being finite, there exists an integer N such that for any non-periodic base-point p and for any j, j′≥ N, p is a base-point ofφj (respectively ofφ− j) if and only if p is a base-point ofφj′ (respectively ofφ− j′
).
We decompose the set of non-periodic base-points ofφinto four sets:
B
++= p| p is a base-point ofφj, and is a base-point ofφ− j for j≥ N ,B
+−= p| p is a base-point ofφj, but is not a base-point ofφ− j for j≥ N ,
B
−+= p| p is not a base-point ofφj, but is a base-point ofφ− j for j≥ N ,
B
Note that
B
+−is the set of persistent base-points ofφand that
B
−+is the set of persistent base-point ofφ−1.This decomposes the set of base-points of φinto five disjoint sets. Two base-points p, p′ ofφwhich are equivalent belong to the same set.
We fix an integer k≥ 2N and compute the number of base-points of φk. Any such base-point being equivalent to a base-point ofφ, we take a base-point p ofφ, and count the number mp,kof base-points ofφk
which are equivalent to p.
If p belongs to
P
, the number of points equivalent to p is less than #(bP
), so mp,k≤ #(bP
).If p is not in
P
, any point equivalent to p is equal to(φi)•(p) for some i, and all are distinct, so we havemp,k= #Ip,k, where Ip,k=
n
i∈ Z
p is not a base-point ofφi, but p is a base-point ofφi+ko.
If p∈
B
++, since p is not a base-point ofφione has−N < i < N, thus mp,k< 2N.A point p∈
B
−−is a base-point ofφi+khence−N < i + k < N and mp,k< 2N.
If p∈
B
−+, the fact that p is not a base-point ofφi implies that−N < i and the fact that p is a base-point
ofφi+k implies that i+ k ≤ N. With these two inequalities one has −N < i ≤ N − k. Since k > 2N, we get mp,k= 0.
If p∈
B
+−, the fact that p is not a base-point of φi implies that i< N and the fact that p is a
base-point of φi+k implies that −N < i + k. This yields −N − k < i < N, so mp,k≤ 2N + k. Conversely, if
i≤ −N and i + k ≥ N, p is not a base-point of φi, but p is a base-point of φi+k (or equivalently i∈ I
p,k),
so mp,k≥ #[N − k,−N] = k − 2N + 1. The two conditions together imply that mp,k− k ∈ [−2N,2N].
Recall that
B
+− is the set of persistent base-points ofφ. The above counting explains that the number of
base-points ofφk, for k big, behaves likeνk, where νis the set of equivalence classes of of persistent base-points ofφ. More precisely, there exist two integers c, d which do not depend on k, such that the total number of base-points ofφk is betweenνk+ c andνk+ d, for any k ≥ 2N. Recalling that µ(φ) = lim
k→+∞ b(φk)
k , where
b(φk) is the number of base-points ofφk, we obtain µ(φ) =ν.
Corollary 3.4. The dynamical number of base-points is an invariant of conjugation: if θ: S99K Z is a
birational map between smooth projective surfaces andφ∈ Bir(S), then µ(φ) = µ(θφθ−1).
In particular, ifφis conjugate to an automorphism of a smooth projective surface, then µ(φ) = 0.
Proof. The mapθfactorises into the blow-up of a finite number of base-points followed by the contraction of a finite number of curves. The number of equivalence classes of persistent base-points ofφandθφθ−1 is thus the same, and we get the result from Proposition3.3.
It is thus clear that µ(φ) = 0 ifφis conjugate to an automorphism of a smooth projective surface. Proposition 3.5. Let S be a smooth projective surface, and let φ∈ Bir(S). The following conditions are
equivalent: (1) µ(φ) = 0;
(2) φis conjugate to an automorphism of a smooth projective surface. Proof. Corollary3.4yields(2) ⇒ (1). It suffices then to show (1) ⇒ (2).
Denote by K the set of points, that belong to S as proper or infinitely near points, which are base-points of
φiandφ− j, for some i, j> 0. Let us prove that K is a finite set. Choosing the smallest possible i, j associated
to p∈ K, the point p is equivalent to (φi−1)•(p), which is a base-point ofφ, and the same holds replacing p with(φk)•(p), for − j < k < i. By this way, we associate a finite number of points of K to each base-point
ofφ. This shows that K is finite. Observing that any point of K is either a proper point of S or is infinitely near, we can blow-up the set K, and obtain a birational morphismπ: bS→ S.
By construction, the birational map bφ∈ Bir(bS) given byπ−1φπ has less base-points thanφ, and satisfies
that no point of bS is a base-point of bφi and bφ− j, for some i, j> 0 (in the notation of [13], the map bφis now algebraically stable). If a(−1)-curve of bS (a smooth curve isomorphic to P1 and having self-intersection
(−1)) is contracted by bφ, we contract it, via a birational morphismη1: bS→ bS1. The map bφ1=η1◦bφ◦ (η1)−1
is again algebraically stable, and we continue the process if a(−1)-curve of bS1 is contracted by bφ1. At the
end, we obtain a birational morphismη=ηk◦ ··· ◦η1: bS→ bSk= eS which conjugates bφtoφk= eφ, and such
that no(−1)-curve of eS is contracted by eφ.
It remains to show that eφis an automorphism of eS. Assuming the converse, we will deduce a contradiction. We write τ1: X1→ eS the blow-up of the base-points of eφ, and write χ1: X1→ eS the morphismχ1= eφτ1,
which is the blow-up of the base-points of eφ−1. Denote by C1⊂ X1 a (−1)-curve contracted byχ1 onto a
base-point p of eφ−1(such a curve exists for each base-point). Because p is a base-point of eφ−1, it is not a base-point of eφrfor any r> 0. The map eφhas no persistent base-point, because µ(φ) = µ(eφ) = 0 (Proposition 3.3
and Corollary3.4). In consequence, there exists k> 1 such that p is not a base-point of eφ−k.
We writeχ2: X2→ X1the blow-up of the base-points of eφ1−kτ1, and denote byτ2: X2→ eS the morphism
τ2= eφ1−kτ1χ2, which is the blow-up of the base-points of(τ1)−1eφk−1; this yields the following commutative
diagram: X2 τ2 ✄✄✄✄ ✄✄✄✄ ✄✄✄✄ ✄✄✄✄ ✄✄ χ2 ❇❇❇ ❇ ❇ ❇ ❇ ❇ X1 χ1 ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ τ1 ~ ~⑥⑥⑥⑥ ⑥⑥⑥⑥ eS e φk−1 / / ❴ ❴ ❴ ❴ ❴ ❴ eS e φ / / ❴ ❴ ❴ ❴ ❴ ❴ ❴ ❴ eS.
Because p is not a base-point of eφ−k, the curve C1⊂ X1 has to be contracted byτ2◦ (χ2)−1, and the curve
C2= (χ2)−1(C1) has self-intersection −1 (otherwise eφ−kwould have a base-point infinitely near to p, and p
itself would then be a base-point).
The curveτ1(C1) is contracted by eφ, and is then not a(−1)-curve. Because it is contracted by eφ1−k, there is
a base-point q of eφ1−kthat is a proper point ofτ1(C1). Sinceτ2is a morphism,τ1χ2blows-up all base-points
of eφk−1, and thus blows-up q. The fact that q is a base-point of eφ1−k, implies that it is not a base-point of eφ, and thus that it is not blown-up byτ1. In consequence, (τ1)−1(q) is a point blown-up byξ2, and which lies
on C1. This is incompatible with the fact that C2= (χ2)−1(C1) has self-intersection −1.
Remark 3.6. In [13, Theorem 0.4] one can find a characterisation of hyperbolic birational mapsφwhich are conjugate to an automorphism of a projective surface. Ifφ∈ Bir(P2) is hyperbolic, we conjugate it to a
birational map of a smooth projective surface S where the action is algebraically stable (this means that the
(φ|H1,1(S))n= (φn)|H1,1(S) for each n); its action on H1,1(S) admits the eigenvalue λ(φ) > 1 with eigenvector
θ+. The mapφis birationally conjugate to an automorphism if and only if(θ+)2= 0.
Proposition 3.5gives another characterisation, for all maps φ∈ Bir(P2) (not only hyperbolic maps),
de-pending only on µ(φ).
Example 3.7. In [2,3,4,5,12], automorphisms with positive entropy are constructed starting from a bira-tional map of P2. In [12] the authors take a birational mapψ∈ Bir(P2), and choose A ∈ Aut(P2) such that
Aψis conjugate to an automorphism of a surface with dynamical degree> 1. The way to find A is exactly to
ensure that Aψhas no persistent base-point (i.e. µ(Aψ) = 0). Let us give an example ([12]): Letφ= Aψbe the birational map given by
A : (x : y : z) 99K αx+ 2(1 −α)y + (2 +α−α2)z : −x + (α+ 1)z : x − 2y + (1 −α)z
withα∈ C \ {0, 1} and
ψ:(x : y : z) 99K (xz2+ y3: yz2: z3).
The mapψ(resp. ψ−1) has five base-points, p= (1 : 0 : 0) and four points infinitely near; we will denote bP1
(resp. bP2) the collection of these points. The automorphism A is chosen such that:
• bP1, A bP2, AψA bP2have distinct supports;
• bP1= (Aψ)2A bP2.
In particular the base-points of φ are non-persistent, so φ is conjugate to an automorphism of a rational surface. More preciselyφis conjugate to an automorphism with positive entropy on P2blown up in bP1, A bP2
and AψA bP2(see [12, Theorem 3.1]).
4. GROWTH OFJONQUIÈRES TWISTS
Lemma 4.1. Let φbe a birational map of P2 which preserves the pencil of lines passing through some
point p0. The set
degφk− k ·µ(φ) 2 k ≥ 0 ⊂ Z is bounded.
In particular, the sequencedegφk k∈Ngrows linearly if and only if µ(φ) > 0 and its growth is given by
µ(φ)
2 ∈
1 2N.
Remark 4.2. Conjugatingφby a map which preserves the pencil does not change the growthdegφk k∈N, but conjugating it by a map which does not preserve the pencil can increase it (see Proposition4.4).
Proof. For any k,φk preserves the pencil of lines passing through p0. It implies that the linear system ofφk
(which is the pull-back of the system of lines of P2byφk) has multiplicity degφk− 1 at p0and has exactly
2(degφk− 1) other base-points, all of multiplicity 1. In particular, degφk= ⌊b(φk)
2 ⌋. The result follows then
directly from Proposition3.3.
Example 4.3. Let us consider the family of birational maps studied in [11] and defined as fα,β:(x : y : z) 99K ((αx+ y)z :βy(x + z) : z(x + z)), α,β∈ C∗.
Any of the fα,βhas three base-points: (1 : 0 : 0), (0 : 1 : 0) and (−1 :α: 1), and preserves the pencil of lines
passing through(1 : 0 : 0). Checking that (−1 :α: 1) is the only one persistent base-point ([11, Theorem 1.6]), the growth ofdeg fαk,β k
∈Nis given by
k
2(see [11, Lemma 1.4]).
Proposition 4.4. Letφ∈ Bir(P2) be a Jonquières twist. There exists an integer a ∈ N such that
lim k→+∞ deg(φk) k = a 2µ(φ) 2 . Moreover, a= 1 if and only ifφpreserves a pencil of lines.
Proof. Sinceφis a Jonquières twist, there existsψ∈ Bir(P2) such that eφ=ψφψ−1 preserves the pencil of
lines passing through some point p∈ P2. Letπ: F1→ P2 be the blow-up of p∈ P2, and let bφ∈ Bir(F1)
be bφ=π−1eφπ. Denote by L
P2 the linear system of lines of P2and byΛthe linear system on F1corresponding to the image byπ−1ψof the system of lines of P2. The degree ofφk is equal to the free intersection of LP2 withφk(LP2), which is the free intersection ofΛwith bφ(Λ).
On F1,Λis linearly equivalent to aL+ b f , where L =π−1(LP2), f is the divisor of a fibre and where a, b∈ N. Note thatφk(L) is the transform on F
1of the linear system of eφ−k, and is thus equal to L+ (dk− 1) f ,
where dk is the degree of eφk (and of eφ−k). The system bφk(Λ) is then linearly equivalent to aL + (a(dk− 1) +
b) f , so the total intersection of bφk(Λ) withΛis a2d
k+ 2ab. Because bφk(Λ) · f = a, each base-point of bφk(Λ)
has at most multiplicity a. By Lemma4.1, lim
k→+∞
dk
k = µ(φ)
2 . The number of base-points ofΛbeing bounded, the free intersection of bφk(Λ) withΛgrows like a2 µ(φ)
2 · k.
It remains to see that if a= 1 thenφpreserves a pencil (the other direction follows from Lemma4.1). If a= 1, one gets f ·Λ= 1. This implies that the free intersection ofφ−1π( f ) withφ−1π(Λ) = L
P2 is 1; so
φ−1π( f ) is a pencil of lines, invariant byφ.
Lemma4.1and the second assertion of Corollary3.4imply the following statement of [13, Theorem 0.2.]
Corollary 4.5. Letφbe a Jonquières twist; thenφis not conjugate to an automorphism. We can also derive the following new results.
Corollary 4.6. Letφbe a Jonquières twist. If φmand φn are conjugate in Bir(P2) for some m, n ∈ Z, then
|m| = |n|.
Proof. The fact that degφk k
∈N grows linearly implies that φ preserves a pencil of rational curves [13,
Theorem 0.2]. In particular φis conjugate to a birational map of P2 which preserves the pencil of lines passing through some fixed point p0. According to Lemma4.1, one finds µ(φ) > 0.
Asφmandφnare conjugate in Bir(P2) one has µ(φm) = µ(φn) (Corollary3.4). Since µ(φk) = |k · µ(φ)| for
any k, we get|m| = |n|.
5. GROWTH OFHALPHEN TWISTS
In Section4(especially Lemma4.1), we described the degree growth of a Jonquières twistφ, and showed that it is given by µ(φ), a birational invariant given by the growth of base-points. For an Halphen twist, the
dynamical number of base-points is trivial, but the growth can also be quantified by an invariant. Recall that an Halphen twistφpreserves an unique pencil of elliptic curves. By [14, Theorem 2 and Proposition 7, page 127], a power ofφpreserves any member of the pencil, and acts on this via a translation.
Proposition 5.1. Letφ∈ Bir(P2) be an Halphen twist.
(1) The set lim k→+∞ deg(ψφkψ−1) k2 ψ∈ Bir(P2)
admits a minimum, which is a positive rational numberκ(φ) ∈ Q. Ifφacts via a translation on each member of its invariant pencil of elliptic curves, thenκ(φ) ∈ 9N.
(2) There exists an integer a≥ 3 such that lim
k→+∞
deg(φk)
k2 =κ(φ) ·
a2
9.
(a) a= 3;
(b) φpreserves an Halphen pencil, i.e. a pencil of(elliptic) curves of degree 3n passing through 9
points with multiplicity n.
Proof. An Halphen twist preserves an unique elliptic fibration, so there exists an elementψin Bir(P2) such
thatφ′=ψφψ−1preserves an Halphen pencil. Denoting byπ: S→ P2the blow-up of the 9 base-points of the
pencil, bφ=π−1φ′πis an automorphism of S, which preserves the elliptic fibration S→ P1given by| − mK S|
for some positive integer m.
Replacing bφby some power if needed, we can assume that bφis a translation on a general fibre. As explained in the proof of [14, Proposition 9, page 132], this yields the existence of an element∆∈ Pic(S) (depending on bφ) with∆· KS= 0 such that the action of bφon Pic(S) is given by
D7→ D − m(D · KS) ·∆+γKS,
whereγis an integer depending on D which can be computed using the self-intersection:
γ= −m
2
2 (D · KS) ·∆
2+ m(D ·∆).
We denote by L the linear system of lines of P2and byΛ=π−1ψ(L) its transform on S. The degree ofφn
is equal to the free intersection of L withφn(L), which is equal to the free intersection ofΛwith bφn(Λ).
The map bφnacts on Pic(S) as
D7→ D − m(D · KS) · (n∆) + −m 2 2 (D · KS) · (n∆) 2+ m(D · (n∆)) KS. This yields Λ· bφn(Λ) = Λ2− m(Λ· K S) · (n∆·Λ) + −m2 2 (Λ· KS)(n∆) 2+ m(nΛ·∆)(K S·Λ) = Λ2+−m2 2 (Λ· KS)2(∆)2 · n2.
The free intersection betweenΛand bφn(Λ) is thus equal to
Λ2+ −m 2 2 (Λ· KS) 2(∆)2 · n2−
∑
k i=1 µi(Λ) · µi(bφn(Λ)),where µi(Λ) and µi(bφn(Λ)) denote the multiplicities of respectively Λand bφn(Λ) at the r base-points ofΛ.
Since bφis an automorphism of S, the contribution given by the base-points is bounded, so we find that lim n→+∞ deg(φn) n2 = m 2(Λ · KS)2· −∆2 2 .
Note thatΛ is the lift byπ−1 of the homaloidal linear system ψ(L), so Λ· (−KS) ≥ 3, and equality holds
if and only ifΛhas no base-point. This shows that the minimum among all homaloidal systems is attained whenΛhas no base-point; we getκ(φ) = 9m2·−∆2
2
and a=Λ· KS.
Let us prove that−∆2is a positive even number. To do this, we take the orthogonal basis L, E1, . . . , E9of
Pic(S), where L is the pull-back of a line of P2 and E
1, . . . , E9are the exceptional divisors associated to the
points blown-up. Writing
∆= dL −
9
∑
i=1
the equality∆·KS= 0 implies that∑ai= 3d. Modulo 2, we have∑(ai)2≡∑ai≡ 3d ≡ d2, so∆2= d2−∑(ai)2
is even. Applying Cauchy-Schwarz to(1, . . . , 1) and (a1, . . . , a9), we get 9∑(ai)2≥ (∑ai)2, and equality holds
only when all ai are equal. This latter would imply that∆is a multiple of KS, and thus that bφacts trivially on
Pic(S). Hence we have 9∑(ai)2> (∑ai)2= (3d)2, which implies that−∆2=∑(ai)2− d2> 0.
The fact that−∆2is a positive even number implies thatκ(φ) = 9m2·−∆2
2
is a positive integer divisible by 9. The equalityκ(φ) = lim
k→+∞
deg(φk)
k2 is equivalent to the fact thatΛhas no base-point, which corresponds to say that ψ−1πis a birational morphism, or equivalently that the unique pencil of elliptic curves invariant
byφis an Halphen pencil.
Corollary 5.2. Letφ∈ Bir(P2) be an Halphen twist. The integerκ(φ) is a birational invariant which satisfies
κ(φm) = m2κ(φ) for any m ∈ Z. In particular, the mapsφnandφmare not conjugate if|m| 6= |n|.
Proof. Is a direct consequence of Proposition5.1.
Let us give an example whereκ(φ) is not an integer.
Example 5.3. LetΛbe the pencil of cubic curves of P2 given byλ(x2y+ z3+ y2z) + µ(x2z+ y3+ yz2) = 0,
(λ: µ) ∈ P1. The pencil is invariant by the automorphismα∈ Aut(P2) defined by (x : y : z) 7→ (ix : −y : z). Denote byπ: S→ P2the blow-up of the base-points ofΛ, which are 7 proper points of P2and 2 infinitely near points. More precisely, the 7 proper points are
p1= (1 : 0 : 0), p2= (i √ 2 : 1 : 1), p3=α(p2) = (− √ 2 :−1 : 1), p4=α(p3) = (−i √ 2 : 1 : 1), p5=α(p4) = ( √ 2 :−1 : 1), p6= (0 : i : 1), p7=α(p6) = (0 : −i : 1).
The last two points are the following: the point p8 is infinitely near to p6, corresponding to the tangent
direction of the line y= iz, and p9 is infinitely near to p7, corresponding to the tangent direction of the line
y= −iz.
The surface S inherits an elliptic fibration S→ P1, and the lift ofα yields an automorphism bα=παπ−1
of S. Denote by Ei∈ Pic(S) the divisor of self-intersection −1 corresponding to the point pi. If i6= 6, 7,
then Eicorresponds to a(−1)-curve of S; and E6, E7correspond to two reducible curves of S.
For any∆∈ Pic(S) satisfying∆· KS= 0, we denote byι∆∈ Aut(S) the automorphism which restricts on a
general fibre C to the translation given by the divisor∆|C. If∆2= −2, the action ofι∆on Pic(S) is given by
(see the proof of Proposition5.1)
D7→ D − (D · KS) ·∆+ (D · (KS+∆)) · KS.
For any automorphismσ∈ Aut(S), one can check thatισ(∆)=σι∆σ−1. In particular, we have
(αιb ∆)4= (αιb ∆bα−1)(bα2ι∆bα−2)(bα3ι∆αb−3)ι∆=ιαb(∆)ιbα2(∆)ιαb3(∆)ι∆=ιbα(∆)+bα2(∆)+bα3(∆)+∆. Because of the action ofαon the points pi, we have
b
α(E6) = E7,bα(E7) = E6,αb(E2) = E3,αb(E3) = E4,αb(E4) = E5,αb(E5) = E2.
We now fix∆∈ Pic(S) to be the divisor E2− E6(that satisfies∆· KS= 0 and∆2= −2), and obtain
b
α(∆) +bα2(∆) +bα3(∆) +∆= E
which has square−8. In particular, the numberκassociated to(bαι∆)4=ι b α(∆)+bα2(∆)+bα3(∆)+∆is −9(ιbα(∆)+bα2(∆)+bα3(∆)+∆) 2 2 = 36.
This shows, by Corollary 5.2, thatκ(φ) = 9
4, whereφis the birational map of P
2 conjugate toαιb
∆ byπ−1, namelyαπι∆π−1.
6. APPLICATIONS
6.1. Birational maps having two conjugate iterates.
Lemma 6.1. Letφdenote a birational map of P2. Assume that φn and φm are conjugate and assume that
|m| 6= |n|. Then,φis an elliptic and satisfiesλ(φ) = 1 and µ(φ) = 0.
Proof. The mapφmis conjugate toφnin Bir(P2) so one getsλ(φ)|m|=λ(φ)|n|and|m| · µ(φ) = |n| · µ(φ). This
yieldsλ(φ) = 1 and µ(φ) = 0.
The fact thatλ(φ) = 1 implies that φis elliptic, or a Jonquières or Halphen twist. The Jonquières and
Halphen cases are impossible (Corollaries4.6and5.2).
Proposition 6.2. Letφdenote a birational map of P2of infinite order. Assume thatφnandφmare conjugate and assume that|m| 6= |n|. Then,φis conjugate to an automorphism of C2of the form(x, y) 7→ (αx, y + 1),
whereα∈ C∗such thatαm+n= 1 orαm−n= 1.
In particular, ifφis conjugate toφnfor any positive integer n, thenφis conjugate to(x, y) 7→ (x,y + 1). Proof. Follows from Corollary2.5, Corollary5.2and Lemma6.1.
6.2. Morphisms of Baumslag-Solitar groups in the Cremona group. For any integers m, n such that mn6= 0, the Baumslag-Solitar group BS(m,n) is defined by the following presentation
BS(m, n) = hr, s|rsmr−1= sni.
Recall that if G is a group, the derived groups of G are
G(0)= G, G(i)=hG(i−1), G(i−1)i= hghg−1h−1|g, h ∈ G(i−1)i for all i ≥ 1, and that G is solvable if there exists an integer N such that G(N)= {id}.
The groups BS(m, n) (resp. the subgroups of finite index of BS(m, n)) are solvable if and only if |m| = 1 or|n| = 1 (see [18, Proposition A.6]).
A group G is said to be residually finite if for any g in G\ {id} there exist a finite group H and a group homomorphism Θ: G→ H such that Θ(g) belongs to H \ {id}. The group BS(m,n) is residually finite if and only if|m| = 1 or |n| = 1 or |m| = |n| (see [17]). Let V be an affine algebraic variety; according to [1] any subgroup of finite index of the automorphisms group of V is residually finite. Therefore if|m| 6= |n| and |m|,|n| 6= 1 there is no embedding of BS(m,n) into the group of polynomial automorphisms of the plane.
There is an other proof using the amalgated structure of the group of polynomial automorphisms of the plane and the fact that BS(m, n) is not solvable ([10, Proposition 2.2]).
Lemma 6.3. Letρbe a homomorphism from BS(m, n) = hr, s|rsmr−1= sni to Bir(P2). Assume that |m|,|n|
and 1 are distinct. Ifρ(s) has infinite order, the image of the subgroup of finite index hr, sm|rsm2r−1= smni of
Proof. By Proposition6.2, one can conjugateρso thatρ(s) : (x, y) 99K (αx, y+1) whereα∈ C∗andαm−n= 1
or αm+n = 1. Denoting respectively by ψ the map x,n my or x−1,n my one has ψρ(s)mψ−1 =ρ(s)n.
Soρ(r) =ψτwhereτcommutes withρ(s)m:(x, y) 7→ (αmx, y + m).
According to Lemma2.8, one then hasτ= (η(x), y + R(x)) for someη∈ PGL(2,C),η(αmx) =αmη(x),
and some R∈ C(x) satisfying R(αmx) = R(x). And one getsρ(r) = η(x)±1,mn(y + R(x)).
The group generated byρ(sm) andρ(r) is thus solvable.
Corollary 6.4. If|m|, |n| and 1 are distinct, there is no embedding of BS(m,n) into the Cremona group.
Proof. Lemma6.3shows that the image of any embedding would be virtually solvable, impossible when|m|,
|n| and 1 are distinct.
6.3. Embeddings of GL(2, Q) into the Cremona group. To simplify the notation, we will denote in this
last section by(φ1(x, y),φ2(x, y)) the rational map (x, y) 99K (φ1(x, y),φ2(x, y)) from C2to C2.
Let us first give examples of embeddings of GL(2, Q) into the Cremona group.
Example 6.5. Let k be an odd integer and letχ: Q∗→ C∗be a homomorphism such that a7→χ(aak2)is injective.
The morphismρfrom GL(2, Q) to the Cremona group given by
ρ a b c d = x·χ(ad − bc) (cy + d)k , ay+ b cy+ d
is an embedding. Note thatρ(GL(2, Q)) is conjugate to a subgroup of automorphisms of the k-th Hirzebruch
surface Fk. Changing k gives then infinitely many non conjugate embeddings in the Cremona group.
Remark 6.6. Taking k= 1 andχthe trivial map, Example6.5yields the embeddingρ: GL(2, Q) → Bir(P2)
given by ρ a b c d = x cy+ d, ay+ b cy+ d ,
which is obviously conjugate (by extending the actions to P2) to the classical embedding
bρ
a b c d
= (ax + by, cx + dy) .
Theorem 6.7. Letρ: GL(2, Q) → Bir(P2) be an embedding of GL(2, Q) into the Cremona group, then up to
conjugationρis one of the embeddings described in Example6.5. Proof. Let us set
tq= 1 q 0 1 & dm,n= m 0 0 n , q, m, n ∈ Q.
Remark that t1is conjugate to tnin GL(2, Q), for any n ∈ Z r {0}; we can then assume, after conjugation,
thatρ(t1) = (x, y + 1) (Proposition6.2). Asρ(t1/n) commutes withρ(t1) there exist Anin PGL(2, C) and Rn
in C(x) such that
ρ(t1/n) = (An(x), y + Rn(x))
(see Lemma2.8). Let us prove now that An(x) = x. Since tn1/n= t1the element Anis of finite order so Anis
con-jugate to someξx whereξis some root of unity. Henceρ(t1/n) is conjugate to (ξx, y + Q(x)) where Q ∈ C(x)
x, y −∑ n−1 i=1iQ(ξix) n
. Since t1/nis conjugate to t1, Proposition2.4implies thatξ= 1, which achieves to show
that An(x) = x. This implies, with equalityρ(t1/n)n=ρ(t1), that Rn(x) = 1/n. We thus have for any q in Q
ρ(tq) = (x, y + q).
From dm,nt1d−1m,n= tm/none gets (using again Lemma2.8) that
ρ(dm,n) = ηm,n(x), m ny+ Rm,n(x) , ηm,n∈ PGL(2,C), Rm,n∈ C(x).
The map(Q∗)2→ PGL(2,C) given by (m,n) 7→η
m,n is a homomorphism, which cannot be injective. There
exists thus one element dm,n with (m, n) 6= (1,1) such that ρ(dm,n) = x,mny+ Rm,n(x). Note that m 6= n
since the centralizers ofρ(dm,n) andρ(t1) are different. Conjugating by
x, y +Rm,n(x)
m/n−1
, we can assume that
ρ(dm,n) = x,mny. From dm,nda,b= da,bdm,n one gets for any a, b in Q
ρ(da,b) = ηa,b(x),a by , ηa,b∈ PGL(2,C).
The homomorphism Q∗→ PGL(2,C) given by a 7→ηa,a is injective, so up to conjugation by an element
of PGL(2, C) we can assume that for any a ∈ Qr{0,1} there existsχa,a∈ Cr{0,1} such thatηa,a(x) =χa,ax.
This implies the existence ofχa,b∈ C∗for any a, b in(Q∗)2, such thatηa,b(x) =χa,bx.
We now compute the image of M=
0 1
−1 0
. Sinceρ(M) commutes with ρ(d2,2) = (χ2,2x, y) where
χ2,2 ∈ C∗ is of infinite order, there exist R ∈ C(y) and ν∈ PGL(2,C) such that ρ(M) = (xR(y),ν(y))
(Lemma2.8). For any a∈ Q∗, equality Mda,1= d1,aM yields
χa,1x· R(ay),ν(ay)= χ1,ax· R(y), 1 aν(y) .
This implies that R(y) =αy−kandν(y) = βy, for someα,β∈ C∗, k∈ Z, i.e.ρ(M) =αx yk,
β
y
. We use now equality(Mt1)3= id: the second component of (ρ(M)ρ(t1))3being
β(y +β+ 1)
(β+ 1)y + 2β+ 1,
we findβ= −1 and compute (ρ(M)ρ(t1))3= (xα3(−1)k, y) soα3= (−1)k. Sinceρ(M)2= (xα2(−1)k, y)
has order 2, we have−α2= (−1)k, thusα= −1 and k is odd.
Writingχ(a) =χa,1for any a∈ Q, the map Q∗→ C∗given by a→χ(a) is a homomorphism, and one gets
ρ(da,1) = (χ(a)x, ay). The group GL(2, Q) is generated by the maps da,1, taand M, so
ρ a b c d = x·χ(ad − bc) (cy + d)k , ay+ b cy+ d , for any a b c d
∈ GL(2,Q). This yields an embedding if and only if the homomorphism from Q∗ to C∗ given by a7→ χ(aak2) is injective.
It remains to observe that k can be chosen to be positive. Indeed, otherwise one conjugates by 1x, yand
